Select the container from the figure (Figure 1) that represents the dilution of a 4 % (m/v) KCl solution to each of the following: Figure1 of 1 There is a diagram showing several containers. One container is filled with 4 percent of mass to volume solution of NaCl. Container 1 is filled with a solution in which volume is two times less than the volume of NaCl solution. Container 2 is filled with a solution of a volume two times larger compared to the NaCl solution. Container 3 is filled with a solution of a volume two times larger than the volume of the solution in container 2. Part A 2 % (m/v) KCl

Prior to writing the chemical formula of magnesium chloride, a question which any student will need to ask is: A.  What are the oxidation numbers of each atom?

A chemical formula can be defined as a scientific notation that is used to show (represent) the type and total number of atoms that constitute a particular chemical molecule or compound, by using the oxidation number, chemical symbols and subscripts.

Hence, a student would need to know the oxidation numbers of each atom when magnesium chloride is formed after a chemical reaction.

In Chemistry, an oxidation number represents the number of electrons gained or lost by an atom of a chemical element during a chemical reaction.

In conclusion, when magnesium chloride is formed, a question which the student will need to ask prior to writing the chemical formula is "what are the oxidation numbers of each atom?"

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Waste is removed from the body in various ways. Kidneys filter out metabolic waste, excess fluid and eliminate it in the form of urine. Lungs exhale CO2 from the deoxygenated blood. The undigested solid waste is egested out from the large intestine. Excretion is the process of removing wastes and excess water from the body. It is one of the major ways the body maintains homeostasis. Organs of excretion make up the excretory system. They include the kidneys, large intestine, liver, skin, and lungs.

Answer:

1.false

2.true

3.true

4.true

5.false

6.false

7.false

8.true

9.false

10.true

Explanation:

I took the test

Considering the definition of percentage by mass and molality, the molality of Na₃PO₄ in the solution is 0.948 [tex]\frac{moles}{kg}[/tex].

The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

[tex]percentage by mass=\frac{mass of solute}{mass of solution}x100[/tex]

Molality is the ratio of the number of moles of any dissolved solute to kilograms of solvent.

The Molality of a solution is determined by the expression:

[tex]molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]

Considering 100 grams as a sample of the solution, then the value of the percentage of concentration given indicates that  13.5 g correspond to Na₃PO₄.

Remember that percent concentration by mass is calculated using the mass of solute and the mass of the solution, which includes both the solute and the solvent. Then:

mass solution= mass solute + mass solvent

100 g= 13.5 g + mass solvent

100 g - 13.5 g= mass solvent

86.5 g= mass solvent

Then, you know:

Then, replacing in the definition of molality:

[tex]molality=\frac{0.082 moles}{0.0865 kg}[/tex]

Solving:

molality= 0.948 [tex]\frac{moles}{kg}[/tex]

Finally, the molality of Na₃PO₄ in the solution is 0.948 [tex]\frac{moles}{kg}[/tex].

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In this case, according to the given information about the oxidation numbers and the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.

Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.

Next, we can write the following [tex]x[/tex], since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:

[tex]Mn^xI_2^-\\\\Mn ^xO_2^{-2}[/tex]

Next, we multiply each anion's oxidation number by the subscript, to obtain the following:

[tex]Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4[/tex]

Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.

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This  problem provides the molar mass and radius of a metal that has an BCC unit cell and the density is required.

Firstly, we consider the formula that relates molar mass and also includes the Avogadro's number and the volume of the unit cell:

[tex]\rho =\frac{Z*M}{V*N_A}[/tex]

Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since BCC is able to hold 2 atoms and M and NA are given, we calculate the volume of the atom in the unit cell given the radius in meters:

[tex]V=a^3=(\frac{4R}{\sqrt{3} } )^3=(\frac{4*1.74x10^{-10}m}{\sqrt{3} } )^3=6.49x10^{-29}m^3[/tex]

And finally the required density in g/cm³:

[tex]\rho =\frac{2*341.1g/mol}{6.49x10^{-29}m^3\frac{m^3}{atom} *6.022x10^{23}\frac{atom}{mol} } =17455257.8g/m^3\\\\\rho=17.5g/cm^3[/tex]

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Answer:

It is an unstable form of a chemical element that releases radiation as it breaks down and becomes more stable.

Answer:

a. the breaking of bonds

Answer:

B

Explanation:

Hey, I am almost positive that the answer to this question is B. The breaking of bonds uses the energy but when they are created it releases energy. I hope this helps, Have a good day!

From the information provided in the question, the second ionization energy of the metal is  578 kJ/mol.

From the question, we have the following information;

Lattice enthalpy of MO(s) = -2383 kJ/mol

Bond dissociation enthalpy of O2(g) = +498 kJ/mol

First electron affinity of O = -141 kJ/mol

Second electron affinity of O = +744 kJ/mol

First ionization energy of M = + 267 kJ/mol

Heat of sublimation of M = + 130 kJ/mol

Standard enthalpy of formation of MO(s) = -307 kJ/mol

Using Hess law of constant heat summation;

ΔHf = ΔHs + BE + ∑IE + ∑EA + U

ΔHs  = Heat of sublimation of metal

ΔHf = Heat of formation MO

BE = Bond energy of O2

∑EA  = sum of electron affinities of Oxygen

∑IE  = Sum of the ionization energies of M

U = Lattice energy of MO

Let the second ionization energy be x

Substituting values;

(-307) = 130 + 498 + (267 + x) + 603 + (-2383)

(-307) = -885 + x

-x =   -885 + 307

-x = -578

x = 578 kJ/mol

The second ionization energy of the metal is  578 kJ/mol.

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Answer:

The answer is A and C.

Explanation:

The reactants in the process of cellular respiration are oxygen and glucose, respectively. It is ATP that serves as the primary product of cellular respiration, with carbon dioxide and water serving as waste products.

Sugar is a glucose.

Oxygen and glucose are the two  substances that are reactants in the chemical reactions of cellular respiration. Therefore, the correct options are options A, C.

Through the process of cellular respiration, organisms mix oxygen with food molecules, directing the chemical energy contained in these substances towards life-sustaining processes while excreting carbon dioxide and water as waste. Foods are broken down by microorganisms that do not require oxygen in a process known as fermentation.

Adenosine triphosphate (ATP), an energy-rich compound that absorbs the chemical energy generated by the decomposition of food molecules then releases it to power other cellular functions, is one goal of the breakdown of foodstuffs. ATP is created when the energy found inside chemical bonds is converted from one form to another. Oxygen and glucose are the two  substances that are reactants in the chemical reactions of cellular respiration.

Therefore, the correct options are options A, C.

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Answer:

increase the number of gas particles

Explanation:

Answer:

Forged parts are often tougher than cast parts. This can be determined by performing tensile tests on various areas on the parts. Additionally, the microstructures of forged and cast parts can be used to determine if a part was forged or cast. The microstructure of a cast part will have a more uniform grain structure.

Explanation:

Answer:

310K

Explanation:

Rearrange PV=nRT to get T=PV/nR

T=(2.0L)(0.75atm)/(0.5mol)(0.08206)

=36.5 or 37

add 273 for K to get

310

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)   ΔH  (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).    

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq)   ΔHsoln    (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq)     ΔHneut     (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.      

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq)    (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq)  → H₂O(l) + KCl(aq)    (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

[tex] \Delta H = \Delta H_{soln} + \Delta H_{neut} [/tex]

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

[tex] \Delta H = \Delta H_{neut} - \Delta H_{soln} [/tex]

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

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I hope it helps you!

Answer:

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. ... After the sample has been applied on the plate, a solvent or solvent mixture (known as the mobile phase) is drawn up the plate via capillary action.

According to the equation given, we have N2 + 3H2 → 2NH3.

Therefore, to determine the moles of NH3 we have to do the following-

1.3 mol H2 × 2 mol NH3 / 3 mol H2 = 0.87 mol

Answer: .87 moles of NH3 are produced from 1.3 moles of H2.

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The number of moles of gas B present in the gas mixture is 2.72 moles.

The given parameters;

The total number of moles of the gases is calculated by applying ideal gas law;

PV = nRT

where;

R is ideal gas constant = 0.0821 L.atm./mol.K

n is the total mole of the gases

[tex]n =\frac{PV}{RT} \\\\n = \frac{8 \times 14}{0.0821 \times 323} \\\\n = 4.22 \ moles[/tex]

The number of moles of gas B is calculated as follows;

[tex]n_B = 4.22 - n_A\\\\n_B = 4.22 - 1.5\\\\n_B = 2.72 \ moles[/tex]

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Explanation:

83.6-24.2= 59.4 which is the change in heat

Answer:

Explanation:

Answer:

i can't understand the question

Answer:

Neon 10,  mass: 20.18

Explanation:

The amount of the sample remaining at 4pm is 0.0625 g

We'll begin by calculating the number of half-lives that has elapsed

Half-life (t½) = 1 hour

Time (t) = 4 hour

n = t / t½

n = 4 / 1

Number of half-lives (n) = 4

Initial amount (N₀) = 1 g

[tex]N = \frac{N_0}{ {2}^{n}} \\ \\ N = \frac{1}{ {2}^{4}} \\ \\ N = \frac{1}{16} \\ \\ N = 0.0625 \: g[/tex]

Thus, the amount remaining at 4pm is 0.0625 g

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Answer:

im pretty sure its 2.7 gms

Explanation:

im not for sure tho

Answer:

this the partial removal of a substance from a solution or mixture by dissolving it in another immiscible solvent in which it is more soluble.

Answer:

44

Explanation:

Answer:

Atoms themselves do not have a smell. Molecules do.

Explanation:

What happens is that particular molecules bind to receptors in our nose, activating nerves, that then send signals to our brain, which we then interpret and perceive as “smells”. Different molecules bind to different receptors, which is why we can perceive a number of different smells

Answer:

B

Explanation:

as temperature rises, the particles gain kinetic energy (they will move faster) so if temperatures stays constant, so will the movement or vibration of the particles

Rutherford proposed the nuclear model of atom. The only one model which can be inferred from this model is nucleus with protons and revolving electrons around.

Rutherford proposed some aspects of atomic structure based on his gold foil experiment. He discovered that the alpha ray is scattering from foil by the repulsion it experienced from the metal.

This results lead to the discovery of positively charged particles in atom latter called as protons. Rutherford proposed that theses protons are located inside the nucleus and the electrons are revolving around the nucleus.

Latter Niels Bohr his student interpreted the nuclear model of Rutherford with quantum mechanics and theory of max planck and he proposed the equations for the determination of energy and momentum of electrons and the radius of atom.

Therefore, no other inferences except the presence of protons and revolving electrons can be obtained from Rutherford's model of atom.

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Answer:

go get the stuff.

Explanation: